Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $k = \dfrac{q^2 - 6q - 40}{q + 10} \times \dfrac{q + 10}{q - 10} $
Answer: First factor the quadratic. $k = \dfrac{(q - 10)(q + 4)}{q + 10} \times \dfrac{q + 10}{q - 10} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (q - 10)(q + 4) \times (q + 10) } { (q + 10) \times (q - 10) } $ $k = \dfrac{ (q - 10)(q + 4)(q + 10)}{ (q + 10)(q - 10)} $ Notice that $(q + 10)$ and $(q - 10)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ \cancel{(q - 10)}(q + 4)(q + 10)}{ (q + 10)\cancel{(q - 10)}} $ We are dividing by $q - 10$ , so $q - 10 \neq 0$ Therefore, $q \neq 10$ $k = \dfrac{ \cancel{(q - 10)}(q + 4)\cancel{(q + 10)}}{ \cancel{(q + 10)}\cancel{(q - 10)}} $ We are dividing by $q + 10$ , so $q + 10 \neq 0$ Therefore, $q \neq -10$ $k = q + 4 ; \space q \neq 10 ; \space q \neq -10 $